Muller Dynamo

[hartiberlin]

http://www.youtube.com/watch?v=9pTRtJ0jDjA

Could the position of the rectifier be important? Looks like it does 

Hmm,
very interesting effect.
So the placement and the right diodes and FWBRs seems to be very important !


Good to know...

The question is,  does the Hall effect have a positive or negative effect
on the energy production in the pickup coils ??

Regards, Stefan.

[gyulasun]

Hi yfree,

Your observation was indicated here too:
http://www.overunity.com/index.php?topic=3842.msg284752#msg284752 

In fact Romerouk already had answered it earlier here:

http://www.overunity.com/index.php?topic=3842.msg284234#msg284234

This is his relevant text on this:
The reason for that is that from the output bridge rectifier I am using one diode to send power back to the battery to keep it charged.When the 20watt bulb is connected the power going back to the battery is reduced resulting that slight voltage drop.
This setup was built for about 25watt load, if the load is increased then all setup must be changed.
The gap from the coils to the rotor must be changed and some other things ...

So he used an extra diode  but this setup was in his first video where there was no looping yet.
Would this answer your observation?

rgds,  Gyula

@All,

I am new to this forum, although an old reader of this forum. I value all the contributions by the members of this forum, especially those by Bolt and EMdevices. I would like to ask a simple question about an observation I made, when I watched the first video by Romero (not self-running). I do not understand the strange behaviour of the battery and would appreciate a help from EMdevices or Bolt or any other member, if appropriate. My observation is as follows: There are two digital multimeters connected to the input. One measures the voltage on the battery and the other measures the total current going into the device. When the device reaches a steady state, without load, the input current is 0.94 A and the battery voltage is measured  at 12.59 V. When the load is connected, soon after, the current consumption decreases to 0.92 A. With the decreased  current draw from the battery, the voltage on the battery should go up (less current draw should result in a smaller voltage drop across the internal resistance and the terminal voltage of the battery should increase) . However, in the video, lower current consumption under load results in voltage drop on battery terminals to 12.28 - 12.30 V. When the load is disconnected, the current draw increases again, but the voltage does not drop, as it normally should, it increases.  Is there a logical explanation to this battery behaviour in this circuit?  Thank you for your contributions.

Best regards to All,

yfree

[altair]

The magnetite I have was sent to me from a group on youtube (ummc.org). I have sent some to another guy to play with and would be more then happy to mail you some to try ( $5 for postage).

So I was not able to get the magnetite rotor to even run as a motor ( very weak attraction). I did spin by hand and had gen coil hooked to scope and didn't see any movement, I tried stacking up to 3 magnets on each side of the coils. I am sure it could work as a motor with some changes to the coils ( bigger core center hole), but for now I have reinstalled the neo rotor.
Peace
rawbush

Thanks for the offer, rawbush. I will let you know if I really need some, but I would prefer to not deplete your stock, as getting more from the UMMC might be difficult, I assume.
When you made your cores with the resin, did you apply a magnetic field during curing, to orient the particles ? I think that's important to get good results.

Regards,

Altair

[chrisC]

Hi yfree,

Your observation was indicated here too:
http://www.overunity.com/index.php?topic=3842.msg284752#msg284752 

In fact Romerouk already had answered it earlier here:

http://www.overunity.com/index.php?topic=3842.msg284234#msg284234

This is his relevant text on this:
The reason for that is that from the output bridge rectifier I am using one diode to send power back to the battery to keep it charged.When the 20watt bulb is connected the power going back to the battery is reduced resulting that slight voltage drop.
This setup was built for about 25watt load, if the load is increased then all setup must be changed.
The gap from the coils to the rotor must be changed and some other things ...

So he used an extra diode  but this setup was in his first video where there was no looping yet.
Would this answer your observation?

rgds,  Gyula

@gyulasun & all

Thanks for the info. Whilst on this subject of the bemf loop back diode, do you know how and where that diode is connected? Also, does anyone know why Romero's device takes in almost 1A without a load?  In the first video, just starting the motor with the battery with no load. That  seemed an awful lot of current for 2 driving coils and some mA to operate a couple of Hall devices? Is this a clue to sort sort of static biasing?

chrisC

[mariuscivic]

Hi
@ plengo
@ k4zep
@ mariuscivic
@ Scratchrobot
@ chrisC
Because my language is not English,
Can not correctly understand the "bucking coils"
Please help me confirm the picture in: A, B, C, D which are "bucking configuration"
Thanks!

Hy Arthus!

In order to get the highest voltage you must use figure B or C.(doesn't matter witch)

IF in midlle of the two coils (figures B,D) would have ONLY one polle, then the uotput would be zero becouse the magnetic field travells in the opposite way in both coils. BUT in our setup we have N and S(in midlle) and this means that the mag field travells in the same way.
It's like puting in series two AA batteries

Hope this helps !